Post on 05-Sep-2020
Time : 3 hrs. M.M. : 300
Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005
Ph.: 011-47623456 Fax : 011-47623472
Answers & Solutions
forforforforfor
JEE (MAIN)-2020 (Phase-1)
Important Instructions :
1. The test is of 3 hours duration.
2. The Test Booklet consists of 75 questions. The maximum marks are 300.
3. There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics
having 25 questions in each part of equal weightage. Each part has two sections.
(i) Section-I : This section contains 20 multiple choice questions which have only one correct answer.
Each question carries 4 marks for correct answer and –1 mark for wrong answer.
(ii) Section-II : This section contains 5 questions. The answer to each of the questions is a
numerical value. Each question carries 4 marks for correct answer and there is no negative marking for
wrong answer.
(Physics, Chemistry and Mathematics)
08/01/2020
Morning
JEE (MAIN)-2020 Phase-1 (8M)
2
PHYSICS
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. When photon of energy 4.0 eV strikes the
surface of a metal A, the ejected
photoelectrons have maximum kinetic energy
TA
eV and de-Broglie wavelength A
.The
maximum kinetic energy of photoelectrons
liberated from another metal B by photon of
energy 4.50 eV is TB
= (TA
– 1.5) eV. If the
de-Broglie waelength of these photoelectrons
B
= 2A, then the work function of metal B is :
(1) 1.5 eV (2) 4 eV
(3) 3 eV (4) 2 eV
Answer (2)
Sol. de-Broglie wavelength (),
h
mv p 2m(KE)
h
2mKE
A AB
B A A
T 1.5K(as given)
K T
Also,
A
B
1
2
On solving TA = 2 eV
KB = T
A – 1.5 = 0.5 eV
Work function of metal B is
B = E
B – K
B
= 4.5 – 0.5 = 4 eV
For A, A
= EA
– TA
= 2 eV
2. The length of a potentiometer wire is 1200 cm
and it carries a current of 60 mA. For a cell of
emf 5 V and internal resistance of 20 , the
null point on it is found to be at 1000 cm. The
resistance of whole wire is :
(1) 80 (2) 100
(3) 60 (4) 120
Answer (2)
Sol.
Potential gradient for the potentiometer wire
‘AB’ is
� �
mvm
AB
dV 60 R
d
AP AP
dV 60 RV 1000mV
d 1200
��
VAP
= 50 R mV
Also, VAP
= 5 V (for balance point at P)
3
5R 100
50 10
3. Effective capacitance of parallel combination
of two capacitors C1 and C
2 is 10 F. When
these capacitors are individually connected to
a voltage source of 1 V, the energy stored in
the capacitor C2 is 4 times that of C
1. If these
capacitors are connected in series, their
effective capacitance will be :
(1) 1.6 F
(2) 3.2 F
(3) 4.2 F
(4) 8.4 F
Answer (1)
Sol. In parallel combination, C1 + C2 = 10 F
When connected across 1 V battery, then
1 1
2 2
U C1 1
U 4 C 4
C2 = 8 F C
1 = 2 F
In series combination,
1 2
equivalent1 2
C C 16C 1.6 F
C C 10
JEE (MAIN)-2020 Phase-1 (8M)
3
4. A particle of mass m is fixed to one end of a
light spring having force constant k and
unstretched length l. The other end is fixed. The
system is given an angular speed about the
fixed end of the spring such that it rotates in a
circle in gravity free space. Then the stretch in
the spring is
(1)
2ml
k m(2)
2
2
ml
k m
(3)
2ml
k m(4)
2
2
ml
k m
Answer (4)
Sol.
m
( + x)�0
kx
At elongated position (x),
Fradial
= mr2
kx = m(� + x)2
2
2
mx
k m
�
5. Boolean relation at the output stage-Y for the
following circuit is :
(1) A B (2) A B
(3) A + B (4) A B
Answer (1)
Sol.A
B5 V
Output (Y)
5 V
Diode OR Gate
Transistor NOT Gate
OR + NOT NOR Gate
Y A B A B
6. Consider a solid sphere of radius R and mass
density
2
0 2
r(r) 1 ,
R 0 < r R. The
minimum density of a liquid in which it will float
is :
(1)0
2
5
(2)0
3
(3)0
5
(4)0
2
3
Answer (1)
Sol. For minimum density of liquid, solid sphere has
to float (completely immersed) in the liquid.
mg = FB (Also V
immersed = V
total)
or �
34dV R
3
�
R 22 3
0 2
0
r 44 1 r dr R
3R
�
R3 5
3
0 20
r r 44 R
3 35R
�
3304 R 2 4
R3 5 3
�02
5
7. Proton with kinetic energy of 1 MeV moves
from south to north. It gets an acceleration of
1012 m/s2 by an applied magnetic field (west to
east). The value of magnetic field :
(Rest mass of proton is 1.6 × 10–27 kg)
(1) 0.071 mT
(2) 0.71 mT
(3) 71 mT
(4) 7.1 mT
Answer (2)
JEE (MAIN)-2020 Phase-1 (8M)
4
Sol.
Magnetic Force F = qvB
� qvBa
m perpendicular to velocity.
Also
62K 2 e 10
vm m
6qvB eB 2 e 10a
m m m
3
19 212 3
27
1.6 1010 2 10 B
1.67 10
31
B 10 T 0.71 mT (approx)2
�
8*. The dimension of stopping potential V0 in
photoelectric effect in units of Planck’s
constant ‘h’, speed of light ‘c’ and Gravitational
constant ‘G’ and ampere A is :
(1) h1/3G2/3c1/3A–1
(2) h2/3c5/3G1/3A–1
(3) h–2/3c–1/3G4/3A–1
(4) h2G3/2c1/3A–1
Answer (No option is correct) Bonus
Sol. Stopping potential (V) hxIyGzCr
Here, h = plank’s constant = [ML2T–1]
I = current = [A]
G = Gravitational constant = [M–1L3T–2]
and c = speed of light = [LT–1]
and V = potential = [ML2T–3A–1]
[ML2T–3A–1] = [ML2T–1]x [A]y [M–1L3T–2]z [LT–1]r
Comparing dimension of M, L, T, A, we get
y = –1, x = 0, z = – 1, r = 5
0 1 1 5v h I G c
9. A thermodynamic cycle xyzx is shown on a
V-T diagram
The P-V diagram that best describes this cycle
is : (Diagrams are shcematic and not to scale)
(1)
(2)
(3)
(4)
Answer (1)
Sol.
From the corresponding V-T graph
Process xy Isobaric expansion,
Process yz Isochoric (Pressure decreases)
Process z – x Isothermal compression
Therefore, corresponding PV graph is
JEE (MAIN)-2020 Phase-1 (8M)
5
10. Consider two solid spheres of radii R1 = 1 m,
R2 = 2 m and masses M
1 and M
2, respectively.
The gravitational field due to sphere and
are shown. The value of 1
2
M
M is
(1)1
6
(2)1
2
(3)2
3
(4)1
3
Answer (1)
Sol. From the diagram
Gravitation field at the surface
2
GmE
r
11 2
1
GmE
r
and 22 2
2
GmE
r
2
1 12
2 1 2
E mr
E r m
2
1
2
m2 2
3 1 m
1
2
m 1
m 6
11. The plot that depicts the behavior of the mean
free time (time between two successive
collisions) for the molecules of an ideal gas, as
a function of temperature (T), qualitatively, is
(Graphs are schematic and not drawn to scale)
(1) (2)
(3) (4)
Answer (2)
Sol. Relaxation time mean free path( )
speed
1
v
or 1
T
Graph between 1
v/sT
is a straight line,
12. The graph which depicts the results of
Rutherford gold foil experiment with -particle is
: Scattering angle
Y : Number of scattered -particles detected
(Plots are schematic and not to scale)
(1) (2)
(3) (4)
Answer (4)
JEE (MAIN)-2020 Phase-1 (8M)
6
Sol. In 1911, Ernest Rutherford publish a formula.
Which indicates that number of particles (Y)
that would be deflected by an angle ‘’ due to
scattering is
( )
4
KY
sin2
where K = constant
Corresponding graph.
13. Three charged particles A, B and C with
charges –4q, 2q and –2q are present on the
circumference of a circle of radius d. The
charged particles A, C and centre O of the
circle formed an equilateral triangle as shown
in figure. Electric field at O along x-direction is
(1) 2
0
3q
d(2) 2
0
3 3q
4 d
(3) 2
0
3q
4 d(4) 2
0
2 3q
d
Answer (1)
Sol.
Electric field due to charge +2q at centre O –
� ��
1 20
1 2q 3 i jE
4 2d
Due to –2q
� ��
2 20
1 2q 3 i jE
4 2d
Due to –4q
� ��
3 20
1 4q 3 i jE
4 2d
Net electric field at point O
� � � ��
0 1 2 3 2
0
3 qE E E E i
d
14. In finding the electric field using Gauss law the
formula
enc
0
qE
A
��
is applicable. In the formula
0
is permittivity of free space, A is the area of
Gaussian surface and qenc is charge enclosed
by the Gaussian surface. This equation can be
used in which of the following situation?
(1) Only when the Gaussian surface is an
equipotential surface and E��
is constant on
the surface.
(2) Only when E��
= constant on the surface.
(3) Only when the Gaussian surface is an
equipotential surface.
(4) For any choice of Gaussian surface.
Answer (1)
Sol. By Gauss law
in
0
QE dA
�����
�
If in
0
Q
EA
�
�
or in
0
Q
E A��
then E || A��
Surface is equipotential too.
JEE (MAIN)-2020 Phase-1 (8M)
7
15. The coordinates of centre of mass of a uniform
flag shaped lamina (thin flat plate) of mass
4 kg. (The coordinates of the same are shown
in figure) are
(0, 3) (2, 3)
(2, 2)(1, 2)
(0, 0) (1, 0)
(1) (0.75 m, 1.75 m) (2) (1.25 m, 1.50 m)
(3) (1 m, 1.75 m) (4) (0.75 m, 0.75 m)
Answer (1)
Sol. (0, 3) (2, 3)
(2, 2)(1, 2)
(0, 0) (1, 0)
(1, 3)
C2
C1
For given Lamina
A1 = 1, C1 = (1.5, 2.5)
A2
= 3, C2 = (0.5, 1.5)
cm
1.5 1.5X 0.75
4
cm
2.5 4.5Y 1.75
4
Coordinate of centre of mass
(0.75, 1.75)
16. At time t = 0 magnetic field of 1000 Gauss is
passing perpendicularly through the area
defined by the closed loop shown in the figure.
If the magnetic field reduces linearly to 500
Gauss, in the next 5 s, then induced EMF in the
loop is
16 cm
2 cm4 cm
(1) 48 V (2) 36 V
(3) 56 V (4) 28 V
Answer (3)
Sol. Using faraday law
Induced EMF d dB
Adt dt
4 2dB 1000 50010 10 T/sec
dt 5
16 cm
2 cm4 cm
Area = ( 16 × 4 – 2 × Area of triangle) cm2
21cm64 2 2 4
2
= 56 × 10–4 m2
induced
= 56 × 10–6 V = 56 V
17. The critical angle of medium for a specific
wavelength, if the medium has relative
permittivity 3 and relative permeability 4
3 for
this wavelength, will be
(1) 60°
(2) 45°
(3) 15°
(4) 30°
Answer (4)
Sol. For relative permittivity = 3, = 30
For relative permeability 4,
3
0
4
3
= 40
0
0 0 v 1
c 2
c
1sin
2
c = 30°
JEE (MAIN)-2020 Phase-1 (8M)
8
18. The magnifying power of a telescope with tube
length 60 cm is 5. What is the focal length of its
eye piece?
(1) 30 cm (2) 10 cm
(3) 20 cm (4) 40 cm
Answer (2)
Sol.
fo
fe
Eye-piece
Objective
For telescope
Tube length (L) = fo + f
e
and magnification (m) e
o
f
f
where fo and f
e are focal length of objective
and eyepiece
fo + f
e = 60
and fe = 5f
o
fo = 50 cm
fe = 10 cm
19. Consider a uniform rod of mass M = 4 m and
length l pivoted about its centre. A mass m
moving with velocity v making angle 4
to
the rod’s long axis collides with one end of the
rod and sticks to it. The angular speed of the
rod-mass system just after the collision is
(1)4 v
7 l(2)
3 v
7 l
(3)3 v
l7 2(4)
3 2 v
7 l
Answer (4)
Sol.
P 45°
v
l
2 2
About point P, angular momentum (L) of thesystem
Linitial
mv l
22
2 2
final
(4m)l mlL
12 4
As Linitial
= Lfinal
3 2v
7 l
20. A leak proof cylinder of length 1 m, made of a
metal which has very low coefficient of
expansion is floating vertically in water at 0°C
such that its height above the water surface is
20 cm. When the temperature of water is
increased to 4°C, the height of the cylinder
above the water surface becomes 21 cm. The
density of water at T = 4°C, relative to the
density at T = 0°C is close to
(1) 1.04
(2) 1.03
(3) 1.26
(4) 1.01
Answer (4)
Sol. Law of floatation
bodyi
liquid
V
V
In given case
4 C1 water
0 C2 water
h
h body
constant
4 C
0 C
80100 201.01
79100 21
SECTION - II
Numerical Value Type Questions: This section
contains 5 questions. The answer to each of the
questions is a numerical value. Each question carries
4 marks for correct answer and there is no negative
marking for wrong answer.
21. Four resistances of 15 , 12 , 4 and 10 respectively in cyclic order to form
Wheatstone’s network. The resistance that is
to be connected in parallel with the resistance
of 10 to balance the network is _________ .
Answer (10)
JEE (MAIN)-2020 Phase-1 (8M)
9
Sol. 12 15
4 10
R
G
Wheatstone bridge balance condition
1 3
2 4
R R
R R
15 15
10R4
10 R
10R 15 4
510 R 12
2R = 10 + R
R = 10
22. A one metre long (both ends open) organ pipe
is kept in a gas that has double the density of
air at STP. Assuming the speed of sound in air
at STP in 300 m/s, the frequency difference
between the fundamental and second
harmonic of this pipe is _______ Hz.
Answer (106)
vair
= 300 m/s
gasair
B 300v 150 2 m/s
2 2
th
n harmonics
nvf
2L (open organ pipe)
gas
1 0
v 150 2f f 75 2 Hz.
2L 2 1
= 106 Hz (approx)
23*. A body A, of mass m = 0.1 kg has an initial
velocity of ˆ3i ms–1. It collides elastically with
another body, B of the same mass which has
an initial velocity of ˆ5 j ms–1. After collision, A
moves with a velocity ˆ ˆv 4 i j�
. The energy
of B after collision is written as x
J10
. The value
of x is ________.
Answer (01) Bonus
Sol. By conservation of momentum
A B A Bu u v v� � � �
(Masses are equal)
Bˆ ˆ ˆ ˆ ˆ ˆv (3i 5 j) (4i 4 j) j i
�
22
B B
1 1KE mv 0.1 2
2 2
1J
10
xJ
10
x = 1
24*. A particle is moving along the x-axis with
its coordinate with time ‘t’ given by x(t) = 10 +
8t – 3t2. Another particle is moving along the
y-axis with its coordinate as a function of time
given by y(t) = 5 – 8t3. At t = 1 s, the speed of
the second particle as measured in the frame
of the first particle is given as v . Then v
(in m/s) is ________.
Answer (580) Bonus
Sol. For particle ‘A’ For particle ‘B’
XA = – 3t2 + 8t + 10 Y
B = 5 – 8t3
A
ˆV (8 6t)i�
2
BˆV 24t j
�
A
ˆa 6i�
B
ˆa 48t j�
at t = 1 sec
A B
ˆ ˆv 2i, v 24 j� �
� � �
B/A A Bˆ ˆv v v 2i 24 j
Speed of B w.r.t. A, �
B/Av 4 576 580
2v 580 m/s
25. A point object in air is in front of the curved
surface of a plano-convex lens. The radius of
curvature of the curved surface is 30 cm and
the refractive index of the lens material is 1.5,
then the focal length of the lens (in cm) is
______.
Answer (60)
Sol. Lens-maker formula 1 2
1 11( 1)
R Rf
for plano-convex lens.
R1 then R
2 = – R
R 30f 60 cm.
1 1.5 1
JEE (MAIN)-2020 Phase-1 (8M)
10
CHEMISTRY
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. The complex that can show fac-and mer-
isomers is
(1) [CoCl2(en)
2] (2) [Co(NH
3)
3(NO
2)
3]
(3) [Pt(NH3)
2Cl
2] (4) [Co(NH
3)
4Cl
2]+
Answer (2)
Sol. [Co(NH3)3(NO
2)3] will show fac and mer isomers
2. As per Hardy-Schulze formulation, the
flocculation values of the following for ferric
hydroxide sol are in the order
(1) AlCl3 > K
3[Fe(CN)
6] > K
2CrO
4 > KBr = KNO
3
(2) K3 [Fe(CN)
6] < K
2CrO
4 < KBr = KNO
3 = AlCl
3
(3) K3[Fe(CN)
6] > AlCl
3 > K
2CrO
4 > KBr > KNO
3
(4) K3[Fe(CN)
6] < K
2CrO
4 < AlCl
3 < KBr < KNO
3
Answer (2)
Sol. Fe(OH)3 is a positive sol. Its coagulation will be
caused by the anion of the electrolyte.
The flocculation value is inversely proportional
to coagulation power or valency of the anion.
The correct order of flocculation value is
K3 [Fe(CN)
6] < K
2CrO
4 < KBr = KNO
3 = AlCl
3
3. The decreasing order of reactivity towards
dehydrohalogenation (E1) reaction of the
following compounds is
(A) Cl (B) Cl
(C)
Cl
(D)
Cl
(1) B > A > D > C
(2) B > D > A > C
(3) B > D > C > A
(4) D > B > C > A
Answer (4)
Sol. Dehydrohalogenation of the given halides by E1
mechanism is decided by the stability of
carbocation formed in the first step. The
correct decreasing order of the given halides
towards dehydrohalogenation by E1 mechanism
is
Cl
Cl
(D) (B)
> >
(C) (A)
Cl
Cl>
4. For the Balmer series in the spectrum of H
atom,
H 2 2
1 2
1 1R ,
n n the correct statements
among (I) to (VI) are
(I) As wavelength decreases, the lines in the
series converge
(II) The integer n1 is equal to 2
(III) The lines of longest wavelength
corresponds to n2 = 3
(IV)The ionization energy of hydrogen can be
calculated from wave number of these lines
(1) (I), (II), (III)
(2) (II), (III), (IV)
(3) (I), (III), (IV)
(4) (I), (II), (IV)
Answer (1)
Sol. In Balmer series of H-atom, the electronic
transitions take place from higher orbits to 2nd
orbit and the longest wavelength will
correspond to transition from 3rd orbit to 2nd
orbit.
n1 = 2 and n
2 = 3 for longest wavelength.
As wavelength decreases the lines in the
Balmer series converge. The correct
statements are (I), (II) and (III).
JEE (MAIN)-2020 Phase-1 (8M)
11
5. The first ionization energy (in kJ/mol) of Na, Mg,
Al and Si respectively, are
(1) 786, 737, 577, 496 (2) 496, 577, 786, 737
(3) 496, 737, 577, 786 (4) 496, 577, 737, 786
Answer (3)
Sol. Ionisation energy of elements belonging to
period III in general increases as we move from
left to right with the exception of Group-2 and
Group-15 elements due to their stable
configuration. The increasing order of first
ionisation energy of the given elements is
Na < Al < Mg < Si
Ionisation energy of the given metals are
Na : 496 kJ/mol ; Al : 577 kJ/mol
Mg : 737 kJ/mol ; Si : 786 kJ/mol
6. The predominant intermolecular forces present
in ethyl acetate, a liquid, are
(1) Dipole-dipole and hydrogen bonding
(2) London dispersion and dipole-dipole
(3) Hydrogen bonding and London dispersion
(4) London dispersion, dipole-dipole and
hydrogen bonding
Answer (2)
Sol. The intermolecular forces present in liquid
ethyl acetate are
(i) Dipole-dipole interaction
(ii) London dispersion
7. A graph of vapour pressure and temperature for
three different liquids X, Y, and Z is shown below
Va
po
ur
pre
ss
ure
(mm
Hg
) 800
500
400
200
0 293 313 333 353
X Y Z
Temp
The following inferences are made
(A) X has higher intermolecular interactions
compared to Y.
(B) X has lower intermolecular interactions
compared to Y
(C) Z has lower intermolecular interactions
compared to Y.
The correct inferences is/are
(1) (B) (2) (C)
(3) (A) and (C) (4) (A)
Answer (1)
Sol. Vapour pressure of a liquid at a given
temperature is inversely proportional to
intermolecular force of attraction. At the same
temperature, vapour pressure of X is higher
than that of Y.
Therefore (X) has lower intermolecular
interactions compared to Y. Statement (B) is
correct.
8. The stoichiometry and solubility product of a
salt with the solubility curve given below is,
respectively
[Y]/mM 3
2
1
[X]/mM
1 2 3
(1) X2Y, 2 × 10–9 M3 (2) XY
2, 4 × 10–9 M3
(3) XY2, 1 × 10–9 M3 (4) XY, 2 × 10–6 M3
Answer (2)
Sol. From the given curve,
if [X] = 1 mM then [Y] = 2 mM
Salt is XY2
Ksp
= [X][Y]2 = (10–3)(2 × 10–3)2 = 4 × 10–9 M3
9. When gypsum is heated to 393 K, it forms
(1) Anhydrous CaSO4
(2) Dead burnt plaster
(3) CaSO4 5H
2O
(4) CaSO4 0.5H
2O
Answer (4)
Sol. Gypsum on heating to 393 K forms plaster of
Paris
393K
4 2 4 2 2CaSO 2H O CaSO 0.5H O 1.5H O
10. Arrange the following compounds in increasing
order of C – OH bond length
methanol, phenol, p-ethoxyphenol
(1) phenol < p-ethoxyphenol < methanol
(2) methanol < phenol < p-ethoxyphenol
(3) methanol < p-ethoxyphenol < phenol
(4) phenol < methanol < p-ethoxyphenol
Answer (1)
JEE (MAIN)-2020 Phase-1 (8M)
12
Sol. C-OH bond length in methanol, phenol and
p-ethoxyphenol is least in phenol due to
resonance and maximum in methanol due to
lack of resonance whereas it will have some
intermediate value in p-ethoxyphenol.
correct increasing order is
phenol < p-ethoxyphenol < methanol
11. The rate of a certain biochemical reaction at
physiological temperature (T) occurs 106 times
faster with enzyme than without. The change in
the activation energy upon adding enzyme is
(1) – 6RT (2) + 6RT
(3) + 6(2.303)RT (4) – 6(2.303)RT
Answer (4)
Sol. The rate constant of a reaction is given by
aE /RT
k Ae
The rate constant in presence of catalyst is
given by
aE /RT
k Ae
a a(E E )/RTk
ek
a a(E E )/RT6
10 e
6 a a(E E )
ln 10RT
a aE E 6(2.303)RT
12. The strength of an aqueous NaOH solution is
most accurately determined by titrating
(Note : consider that an appropriate indicator is
used)
(1) Aq. NaOH in a pipette and aqueous oxalic
acid in a burette
(2) Aq. NaOH in a burette and aqueous oxalic
acid in a conical flask
(3) Aq. NaOH in a burette and concentrated
H2SO
4 in a conical flask
(4) Aq. NaOH in a volumetric flask and
concentrated H2SO
4 in a conical flask
Answer (2)
Sol. In the titration of acid with a strong base like
NaOH, the oxalic acid is taken in a conical flask
and NaOH is taken in a burette.
13. The most suitable reagent for the given
conversion is
CONH2
HO C2
CN
C ==O
CH3
?
CONH2
HOH C2
CN
COCH3
(1) LiAIH4
(2) NaBH4
(3) H2/Pd (4) B
2H
6
Answer (4)
Sol. Diborane selectively reduces carboxylic acid to
alcohol in preference to other functional groups
like amide, carbonyl group and cyanide group.
14. Which of the following statement is not true for
glucose?
(1) Glucose reacts with hydroxylamine to form
oxime
(2) The pentaacetate of glucose does not react
with hydroxylamine to give oxime
(3) Glucose exists in two crystalline forms
and (4) Glucose gives Schiff’s test for aldehyde
Answer (4)
Sol. Glucose exists is two anomeric forms and .
It forms oxime with NH2OH and its
pentaacetate does not react with NH2OH
because its anomeric OH group is converted
into acetate group. But glucose does not give
Schiff test for aldehyde
15. The third ionization enthalphy is minimum for
(1) Mn (2) Fe
(3) Co (4) Ni
Answer (2)
Sol. The electronic configurations of the given
metals and in their +3 state are :
Mn : 3d 5 4s2 Mn3+ : 3d 4
Fe : 3d 6 4s2 Fe3+ : 3d 5
Co : 3d7 4s2 Co3+ : 3d 6
Ni : 3d 8 4s2 Ni3+ : 3d 7
Since Fe3+ has stable configuration of 3d5, the
third ionization energy of Fe is minimum.
JEE (MAIN)-2020 Phase-1 (8M)
13
16. Among the gases (a) - (e), the gases that cause
greenhouse effect are
(a) CO2
(b) H2O
(c) CFCs (d) O2
(e) O3
(1) (a), (b), (c) and (d)
(2) (a), (c), (d) and (e)
(3) (a), (b), (c) and (e)
(4) (a) and (d)
Answer (3)
Sol. Green house gases = CO2 , H
2O (Vapour), CFCs
and O3
17. The major product of the following reaction is
OHH C3
H C3H C
3
dil. H SO2 4
(1)
CH3
CH3
OHH C3
(2)
H C3
H C3
OH
OH
CH3
OH
(3)
H C3
H C3
OHOH
CH3
HO
(4)
CH3
OH
CH3H C3
Answer (4)
Sol.
+
H O2
–H+
OH
OH
dil H SO2 4
– H O2CH2
+
18. A flask contains a mixture of isohexane and
3-methylpentane. One of the liquids boils at
63°C while the other boils at 60°C. What is the
best way to separate the two liquids and which
one will be distilled out first?
(1) Simple distillation, isohexane
(2) Fractional distillation, isohexane
(3) Simple distillation, 3-methylpentane
(4) Fractional distillation, 3-methylpentane
Answer (2)
Sol. Two volatile liquids having their boiling points
close to each other can be separated by
fractional distillation. Out of the two given
liquids, isohexane has lower boiling point and
hence will distill out first.
19. The major products A and B in the following
reactions are
CN Peroxide
Heat[A]
[A] + B
(1)CN
ACN
and B
(2)CN
and B
CN
A
(3)CN CN
and B A
(4)CN
CNA and B
Answer (4)
Sol. Peroxide generates a radical that abstracts
H-atom from the C-atom adjacent to CN group
to give more stable radical
CN Peroxide
CN
[A]
[A] attacks 1-pentene to give 2° radical that
picks up H-atom to give [B]
CN CN+
[B]
JEE (MAIN)-2020 Phase-1 (8M)
14
20. The number of bonds between sulphur and
oxygen atoms in 28
2S O and the number of
bonds between sulphur and sulphur atoms in
rhombic sulphur, respectively, are
(1) 4 and 6 (2) 8 and 8
(3) 8 and 6 (4) 4 and 8
Answer (2)
Sol. S O2 8
2– – –
O — S — O — O — S — O
O
O
O
O
Number of S — O bonds in 28
2S O is 8
S8
S
S
S
S
S
SS
S
Number of S — S bonds in rhombic sulphur is 8.
SECTION - II
Numerical Value Type Questions: This section
contains 5 questions. The answer to each of the
questions is a numerical value. Each question carries
4 marks for correct answer and there is no negative
marking for wrong answer.
21. The magnitude of work done by a gas that
undergoes reversible expansion along the path
ABC shown in the figure is _____________.
4 6 8 10 12
4
6
8
10B
C
(2, 2)
Pressure (Pa)
Volume
(m )3
A
Answer (48.00)
Sol. Work done by a gas that undergoes a reversible
expansion along the path ABC is given by
8 12
B
C
2
P (Pa)
Volume(m )
3
A
2
8
1W = 6×6 + × 4 ×6
2 = 48.00 J
22. The volume (in mL) of 0.125 M AgNO3
required
to quantitatively precipitate chloride ions in
0.3 g of [Co(NH3)
6]CI
3 is _____________.
M[Co(NH3)6]CI
3 = 267.46 g/mol
MAgNO3 = 169.87 g/mol
Answer (26.92)
Sol. Number of moles of the given complex,
0.3=267.46
Number of moles of CI– ions = 0.3× 3
267.46
Moles of Ag+ ions needed to ppt. CI– = 0.3× 3
267.46
Let the volume of 0.125 M AgNO3 needed by V ml.
0.125×V 0.3×3=
1000 267.46
V = 26.92 ml
23. The number of chiral centres in penicillin is
_____________.
Answer (3.00)
Sol. General structure of pencillin is
R — C — NH
O
O
H H
* *
N *
S CH3
CH3
COOH
H
The number of chiral centres in pencillin is 3.
24. What would be the electrode potential for the
given half cell reaction at pH = 5? _________.
– 0
2 2 red2H O O + 4H + 4e ; E = 1.23 V
(R = 8.314 J mol–1K–1; Temp = 298 K; oxygen
under std. atm. pressure of 1 bar)
Answer (–0.93)
Sol. The given half cell reaction is
+ –
2 22H O O + 4H + 4e
ox
E = –1.23 V
4+°ox ox 2
0.0591E =E – log Po H
4
JEE (MAIN)-2020 Phase-1 (8M)
15
At PO2 = 1 bar and [H+] = 10–5 M
–20
ox
0.0591E = –1.23 – log10
4
= –1.23 + 0.2955 = –0.93 V
25. Ferrous sulphate heptahydrate is used to fortify
foods with iron. The amount (in grams) of the
salt required to achieve 10 ppm of iron in
100 kg of wheat is_____________.
Atomic weight : Fe = 55.85; S = 32.00; O = 16.00
Answer (4.97)
Sol. Mass of iron needed in 100 kg wheat = 5
6
10×10
10
= 1.0 gm
Molecular mass of FeSO4.7H
2O is 277.85
55.85 gm iron is present in 277.85 gm of salt
1 gm iron is present in 277.85
= 4.97 gm.55.85
�����